Why the Feasible Domain Comes First
In an optimization problem, a formula alone does not say which values of its variable are allowed. The allowed values come from the situation being modeled: lengths may need to be positive, capacity may impose an upper limit, and a boundary configuration may be allowed or may no longer represent the object in the problem.
Finding the feasible domain means identifying exactly which values satisfy all the constraints and describe valid configurations. This must be done before using calculus. Otherwise, a derivative calculation might include values that have no physical meaning, or a boundary case might be missed because its endpoint was excluded without justification.
Keep two questions separate. First, for which inputs can the formula be evaluated? Second, which inputs describe configurations permitted by the problem? The algebraic domain answers the first question; the feasible domain answers the second. They need not be the same.
Definitions and Endpoint Language
An inequality’s direction and equality case determine whether a constraint includes its boundary. For example, \(x\geq 2\) permits \(x=2\), while \(x>2\) does not. When a one-variable feasible domain is an interval, we use a square bracket at an included endpoint and a parenthesis at an excluded endpoint.
| Constraint | Endpoint interpretation |
|---|---|
| \(a\leq x\leq b\) | Both endpoints are included: \([a,b]\). |
| \(a<x<b\) | Neither endpoint is included: \((a,b)\). |
| \(a\leq x<b\) | The left endpoint is included, and the right is excluded: \([a,b)\). |
| \(a<x\leq b\) | The left endpoint is excluded, and the right is included: \((a,b]\). |
Physical language must be interpreted carefully. “At most” usually permits equality; “less than” does not. A length described as positive must be strictly greater than zero. A length described as nonnegative may equal zero, although that value might describe a degenerate object rather than a genuine one.
Translate Each Constraint, Then Intersect
When the model has been reduced to one variable, each condition restricts that variable to a set of allowed values. The feasible domain is the intersection of those sets: a value is feasible only if it satisfies every condition at once.
For interval constraints, the intersection can be found by keeping the most restrictive lower bound and the most restrictive upper bound. The endpoint is included only if every constraint that sets that bound allows equality and the configuration remains valid there.
Worked Example: A Rectangle with Positive Side Lengths
Worked Example: A Rectangle with Positive Side Lengths
A rectangle has perimeter \(24\) meters. Let its width be \(x\) meters and its length be \(y\) meters. Determine the feasible domain for \(x\) and the domain of the area formula.
The perimeter constraint is
Divide by \(2\):
Thus \(y=12-x\). A genuine rectangle must have positive side lengths, so
The second inequality gives \(x<12\). Combining both conditions, the feasible domain is
The area formula is \(A=xy\), so after substitution,
This polynomial can be evaluated algebraically at \(x=0\) and \(x=12\), but neither value is feasible. At \(x=0\), the width is zero and the length is \(12\); at \(x=12\), the width is \(12\) and the length is zero. In either case, the configuration is a collapsed shape, not a rectangle with positive side lengths.
Check an interior value, \(x=5\). Then \(y=12-5=7\), so both lengths are positive and \(2(5)+2(7)=10+14=24\). The area is \(A(5)=5(7)=35\) square meters. Substitution verifies both the constraint and the objective formula for this feasible configuration.
This example illustrates an important distinction: the formula’s algebraic domain includes the endpoints, but the physical feasible domain does not. The zero area at an endpoint is a value of the extended formula, not the area of a feasible rectangle under the stated positive-length requirement.
Worked Example: Closed Constraints and Included Endpoints
Worked Example: Closed Constraints and Included Endpoints
A rectangular display has perimeter \(24\) meters. Its width \(x\) must be at least \(2\) meters, its length \(y\) must be at least \(3\) meters, and its width may not exceed \(7\) meters. Find the feasible interval for \(x\), then evaluate the area at both endpoints.
As before, the perimeter constraint gives \(x+y=12\), so \(y=12-x\). Translate the three size constraints:
The length condition gives \(12-x\geq 3\), hence \(-x\geq -9\), or \(x\leq 9\). The combined conditions are therefore
The most restrictive upper bound is \(x\leq 7\). Since all three original constraints allow equality at their boundaries, the feasible interval is
The area is \(A(x)=x(12-x)\). At the left endpoint, \(x=2\), so \(y=12-2=10\), and
Check the constraints: \(2\geq2\), \(10\geq3\), and \(2\leq7\). Thus the endpoint is feasible. At the right endpoint, \(x=7\), so \(y=12-7=5\), and
Here \(7\geq2\), \(5\geq3\), and \(7\leq7\), so this endpoint is feasible as well. Both endpoint values are included because the constraints permit equality and the resulting rectangles have positive side lengths. The calculations check the objective values by substituting the corresponding dimensions into the area formula.
Notice why an endpoint should not be included merely because it comes from a convenient algebraic bound. It is included here because the inequalities allow equality and the resulting configurations satisfy the physical requirements.
Worked Example: A Boundary Where the Model Breaks Down
Worked Example: A Fixed-Volume Cylinder
A cylinder must have volume \(100\text{ cm}^3\). Let its radius be \(r\) centimeters and its height be \(h\) centimeters. Find the feasible domain for \(r\), and explain why the boundary value \(r=0\) cannot be included.
The volume constraint is \(V=\pi r^2h\). Setting \(V=100\) and solving for \(h\) gives
A cylinder with positive volume must have \(r>0\) and \(h>0\). For every \(r>0\), the formula \(h=100/(\pi r^2)\) gives a positive, finite height, so each such radius produces a cylinder satisfying the volume constraint. The feasible domain is therefore
At \(r=0\), the expression for height has denominator \(\pi(0)^2=0\), so it is undefined. More fundamentally, substituting \(r=0\) into the original volume equation gives \(V=\pi(0)^2h=0\) for every finite \(h\), not the required \(100\text{ cm}^3\). Thus \(r=0\) is not feasible, and there is no finite-height cylinder at that boundary satisfying the model.
For a check, choose \(r=1\text{ cm}\). The constraint requires \(h=100/(\pi(1)^2)=100/\pi\text{ cm}\), which is positive and finite. Substituting back gives \(V=\pi(1)^2(100/\pi)=100\text{ cm}^3\). This confirms that \(r=1\) is feasible. As \(r\) approaches zero through positive values, the required height grows without bound; the endpoint is not a hidden cylinder that can be added to the domain.
Boundary Values and Model Changes
A boundary may represent an ordinary permitted case, a degenerate case, or a situation where the model no longer applies. These possibilities should not be confused.
- Included physical boundary: A constraint such as \(x\leq 7\) may allow \(x=7\). If the configuration remains valid there, the endpoint is closed.
- Excluded degeneracy: A zero side length may satisfy an algebraic equation but fail a requirement that an object have positive dimensions. The endpoint is open.
- Breakdown of the model: A formula may become undefined at a boundary, or the original constraint may become impossible there. The fixed-volume cylinder at \(r=0\) is an example.
Sometimes the boundary has a meaningful interpretation, but it is governed by a different description from the one used in the interior. For example, a device may switch off at zero output, or a geometric object may collapse into a lower-dimensional shape. In such a case, do not assume that the interior formula continues to describe the boundary. Check the original situation and constraints there. If a different rule applies, the boundary must be interpreted using that rule.
There is also a useful distinction between an endpoint value and a limiting value. If an endpoint is excluded, values of the objective near it may approach a finite number, but that number is not automatically attained by a feasible configuration. An optimization problem asks for values produced by feasible inputs; a limit at an excluded input is not itself a feasible output unless the boundary is separately allowed and the model defines the objective there.
A Reliable Domain-First Procedure
State what it measures and include its units when appropriate.
Include equations, lower bounds, upper bounds, and requirements such as positive dimensions.
Keep equality cases when the original condition permits equality; preserve strict inequalities when it does not.
A value that violates even one constraint is not feasible.
Decide whether the object still exists, whether the constraints hold, and whether the objective model is defined there.
This process is not a calculus test for optimality. It establishes where an optimization question is meaningful. Once the feasible domain is correct, later analysis can be restricted to that domain.
Common Mistake: Treating Every Algebraic Endpoint as Feasible
Suppose a rectangle has perimeter \(24\), so \(y=12-x\), and its area formula is \(A(x)=x(12-x)\). A common mistake is to write \(0\leq x\leq12\) simply because these values make both side lengths nonnegative.
That calculation silently changes the requirement from a rectangle with positive side lengths to a shape that may have a zero side. At \(x=0\), the other side is \(12\), and at \(x=12\), the other side is \(0\). Neither configuration is a rectangle with two positive side lengths. The correction is to apply the stated physical condition:
The polynomial \(A(x)\) still has algebraic values at \(0\) and \(12\), but those values do not make the endpoints feasible. Always check the original constraints and the meaning of the object, rather than relying only on the simplified formula.
Check Your Understanding
For each question, explain both the inequality reasoning and the physical interpretation of any boundary.
- A length \(x\) must be at least \(4\) meters and strictly less than \(10\) meters. Write its feasible interval and state which endpoint is excluded.
- A rectangle has perimeter \(30\) centimeters, with width \(x\) and positive length \(y\). Use the perimeter equation to find the feasible domain for \(x\).
- A one-variable model requires \(x\geq 1\), \(x\leq 8\), and \(x\leq 6\). Find the intersection and identify whether its endpoints are included, assuming the physical model is valid at both.
- Why is a value that makes an objective formula algebraically defined not necessarily feasible?
- For a fixed positive cylinder volume, why can the radius not equal zero even if the volume formula \(\pi r^2h\) can be written at \(r=0\) for finite \(h\)?
